Exact Solutions of the Ising ModelBachelor Degree Project 15 c

Author: Ludwig Ridderstolpe

Supervisor: Giuseppe Dibitetto

Subject reader: Ulf Danielsson

Department of Physics and Astronomy

Division of Theoretical Physics

Uppsala University

September 7, 2017

Abstract

This report presents the general Ising model and its basic assumptions. This study

aims to, from diagonalization of the Transfer Matrix, obtain the Helmholtz free

energy and the exclusion of a phase transition for the one-dimensional Ising model

under an external magnetic field. Furthermore from establishing the commutation

relations of the Transfer matrices and using the Kramers–Wannier duality one finds

the free energy and the presence of a phase transition for the square-lattice Ising

model.

Sammanfattning

Den här rapporten presenterar den generella Ising modellen och dess antaganden.

Studien syftar till att fr̊an diagonalisering av överföringsmatrisen erh̊alla Helmholtz

fria energi och utesluta fasöverg̊angar för den endimensionella Ising modellen under

ett externt magnetiskt fält. Ytterligare fr̊an att etablera kommutationsrelationer av

överföringsmatriserna och av Kramers–Wannier dualitet finner vi den fria energin

och en fasöverg̊ang för den tv̊adimensionella Ising modellen.

Contents

1 Introduction 3

2 Background Statistical Mechanics 5

2.1 The Partition Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2 Helmholtz Free Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.3 Magnetization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.4 Phase transitions and critical points . . . . . . . . . . . . . . . . . . . . . 6

3 The General Ising Model 8

3.1 Hamiltonian of the Ising model . . . . . . . . . . . . . . . . . . . . . . . 8

3.2 Exactly solvable models . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

4 The One-dimensional Ising Model 10

4.1 The Partition Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4.2 The Transfer Matrix V . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

4.3 Free Energy and Phase Transition . . . . . . . . . . . . . . . . . . . . . . 13

5 The Square-lattice Ising Model 14

5.1 Partition function Square-lattice Ising model . . . . . . . . . . . . . . . . 14

5.2 The Transfer Matrices V and W . . . . . . . . . . . . . . . . . . . . . . . 15

5.3 Star-Triangle Transformation and Commutation Relation . . . . . . . . . 17

5.4 Inversion of the Transfer Matrices . . . . . . . . . . . . . . . . . . . . . . 21

5.5 The Negation Operator R . . . . . . . . . . . . . . . . . . . . . . . . . . 23

5.6 The Shift Operator C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

5.7 Alternate Form of the Transfer Matrices . . . . . . . . . . . . . . . . . . 26

5.8 Functional Relation of the Eigenvalues . . . . . . . . . . . . . . . . . . . 27

5.9 Parametrization of K and L . . . . . . . . . . . . . . . . . . . . . . . . . 29

5.10 Parametrization with Elliptic Functions . . . . . . . . . . . . . . . . . . . 29

5.11 The form of Λ(u) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

5.12 Zeros of Λ(u) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

5.13 Zeros of the RHS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

5.14 Zeros of the LHS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

5.15 General Expression for the Eigenvalues . . . . . . . . . . . . . . . . . . . 38

5.16 Analyticity and validity for k = 1 and k > 1 . . . . . . . . . . . . . . . . 39

5.17 Maximum Eigenvalue and the Free Energy . . . . . . . . . . . . . . . . . 40

5.18 The critical temperature and phase transition . . . . . . . . . . . . . . . 42

6 Discussion 43

References 45

Appendix A 46

Appendix B 48

Appendix C 50

1 Introduction

The Ising model was introduced by Wilhelm Lenz (1920) and assumes particles in a

crystal structure that may without restriction rotate freely around a fixed lattice point

[1]. The model in its entirety arises from expressing such a system’s Hamiltonian in

two parts: one as the energy contribution from particle to particle interaction and one

as the energy contribution from constraints on the system. Developing Lenz proposal

with Statistical mechanics, the Ising model is completely characterized by Helmholtz free

energy and by exactly calculating the free energy is regarded as the solution of the model.

The field of Statistical mechanics is the study of summarizing microscopic dynamics

of a system to a macroscopic description of the system’s thermodynamic equilibrium

properties [2]. Additionally statistical mechanics offers a formalism such that through

the partition function one finds the free energy and from the free energy one can derive

the magnetization and phase transitions.

Lenz (1922) gave the proposed model to his student Ernst Ising. Later Ising (1925)

published a summarization of a exact calculation of the free energy for the one-dimensional

case with an external field and further proved for that case it demonstrates no phase

transitions at any temperature. Firstly the model was regarded as an oversimplification

with no practical applications, Ising followed with making the claim that there is no phase

transitions in higher dimensional models and later abandon the field of physics altogether

[1]. Despite this due to a growing demand of a deeper understanding of ferromagnetism

the Ising model attracted attention and Rudolf Peierls (1936) published a paper where

he claimed in contrary to Ising’s statement that the Ising model in two and three di-

mensions demonstrates phase-transitions [3]. Hendrik Kramers and Gregory Wannier

(1941) located the critical temperature for the two-dimensional Ising model by relating

the low and high temperature expansion with a transformation (Kramers-Wannier Du-

ality). The big breakthrough came when Lars Onsager (1942) announced his analytical

solutions to the Square-lattice Ising model. There has yet been presented any exact

solution to Square-lattice model with an external field or for higher-dimensional models.

It is of great interest to understand ferromagnetic system’s. The Ising model being

the simplest system in we can observe phase transitions analytically [4], it is a power-

ful model which explains dynamics behind ferromagnets spontaneous magnetization and

offers method of determining phase transitions and critical temperatures.

The aim of this thesis is to study the Ising model to obtain the Helmholtz free energy

and magnetic phase transitions for ferromagnetic systems. By firstly outline the proper-

ties and assumptions of the general Ising model and establish its validity as a description

of ferromagnets. Then with the transfer matrix method calculate the free energy of the

one-dimensional Ising model with an external magnetic field and confirm the exclusion

of a phase transition. Lastly from the commuting transfer matrix method and operators

3

calculate the free energy of the Square-lattice Ising model and confirm the existence of

phase transitions and the models critical temperature with Kramer-Wannier duality.

The content of this report are as follows. In chapter 1 a brief review of necessary

concept of statistical mechanics such as: the partition function, the free energy and phase

transitions. In chapter 2 the Hamiltonian of the General Ising model is formulated. In

chapter 3 the solution of the One-dimensional Ising model and review of the transfer

matrix method. In chapter 4 solution of the Square-lattice Ising model, review of the

commuting transfer matrix method. Appendices explain Kramers–Wannier duality and

necessary complex analysis for the solution of the Square-lattice Ising model.

4

2 Background Statistical Mechanics

In this chapter we will review basic statistical mechanical concepts necessary to under-

stand the development of the solutions for the Ising model. We start with firstly covering

the partition function and the Helmholtz free energy, together with some comments about

their significance. Lastly we explain the concept of magnetization and phase transitions

in a system.

2.1 The Partition Function

Consider a system made up of microstates labeled r with the energy of each microstate

Er, then the probability of the system being in a state r is given by pr such that

pr =1

Ze−βEr (2.1)

where β = kBT , kB is the Boltzmann constant, T is the system’s temperature and e−βEr

is the Boltzmann weight. This probability distribution is called the Boltzman-distribution.

Where the partition function Z normalizes the Boltzmann distribution with:

Z =∑r

e−βEr (2.2)

The partition function expresses all accessible states of the system [5], but its significance

is how it can relate the system microscopic properties that constitute the energy of each

microstate to macroscopic thermodynamic quantities.

2.2 Helmholtz Free Energy

For a system held at constant temperature and volume, from the Legendre transform

of the first law of thermodynamic one get the thermodynamic potential F called the

Helmholtz free energy:

F = −kBT lnZ (2.3)

The Helmholtz free energy can be compared to the entropy S which for a closed

adiabatic system attains it maximum when the system is at equilibrium. For a system in

contact to a heath bath the Helmholtz free energy attains its minimum at equilibrium.

This is a powerful property of the free energy and it is regarded as state function [6].

2.3 Magnetization

The magnetization M is a macroscopic quantity which describes the alignment of particles

magnetic moment ~µ in a certain direction. The mean magnetization for a system of N

5

particles labeled i becomes then

〈M〉 = µN∑i

~si (2.4)

where ~si is the direction of the magnetic moment for a particle i. When a system’s

magnetic moment align such that the mean magnetization is of macroscopic size the

systems is said to be ferromagnetic [7].

One can introduce a constraint to the system in the form of a magnetic field ~H, then

the interaction energy between the magnetic moment of particle and the magnetic field

is:

E = −~µ · ~H (2.5)

Moreover the magnetization M can be calculated from the Helmholtz free energy by:

M = − ∂F∂H

(2.6)

This demonstrates the need for the free energy when considering mainly solids and the

study of their magnetization.

2.4 Phase transitions and critical points

A system is said to be in a thermodynamic phase when considered physical properties

change continuously, from that a phase transition is then associated with a discontinuous

change of the physical property [6].

Consider a ferromagnet magnetized by an magnetic field H, for 0� H a large mag-netization M will be induced in the directions of the external field. But if we reverse the

field 0� H a large but opposite magnetization M will be induced. Further we note if welet the magnetic field decrease to 0+ or 0− we will still measure a non-zero spontaneous

magnetization M0 with directions dependenent on the previous field..

This implies a discontinuity around H = 0 as shown in fig. (1.a). As we discussed

above this can be regarded as a phase transition.

HHH

MMM

a. b. c.

M0 _

_-M0

Figure 1: Graphs of M(H, T) for (a) T < Tc, (b) T = Tc, (c) T > Tc. The pictures is

reconstruction from [8, P. 2].

6

Moreover M is found to be dependent on the temperature, increases in T will result

in a lower M0. For a critical temperature Tc we have that M0 vanishes fig. (1.b) and

M(H, T ) is continuous function with an infinite slope at H = 0. For temprature T > Tc

we find that M(H, T ) becomes analytical everywhere fig. (1.c) [8].

7

3 The General Ising Model

In this chapter we will cover an introduction to the General Ising model and its main

assumptions. Will determine the form of the Hamiltonian, establish a connection to

ferromagnetism and discuss the concept of exactly solvable models.

3.1 Hamiltonian of the Ising model

The Ising model is a discrete mathematical description of particles, where the particle’s

magnetic moment is independent and fixed to lattice configuration of a finite number

of sites. We restrict the magnetic moment for all particles to the same direction and

allow they be parallel or antiparallel. Let the total number sites to be N , labeled i =

{1, 2, 3, ...} and to each site assign a spin variable σi = ±1 indicating ”upwards” (+1) or”downwards” (-1) magnetic moment [8].

4321

8765

Figure 2: Example of a 2D lattice configuration, with arrows indicating directions of

magnetic moment

Each spin configuration corresponds to one microstate where then the Hamiltonian of

the system depends on the total spin configuration. The spin configuration σ is the set

of the direction for the spins for all sites

σ = {σ1, ..., σN} (3.1)

then we have in total 2N number of possible spin configurations. The Hamiltonian is

assumed to be

E(σ) = E0(σ) + E1(σ) (3.2)

with E0 the energy contribution from intermolecular interactions between sites and if the

system is under the presence of an external magnetic field then E1 the energy contribution

from spin interaction with the external field. Lastly we note that flipping the direction

of all magnetic moments i.e. negating the spin configuration must clearly not affect

intermolecular energy, giving us:

E0(σ) = E0(−σ) (3.3)

For the intermolecular interaction we assume that only near-neighbouring sites inter-

acts. Meaning of two sites with spin variables σi and σj is aligned (±) and (±) we have a

8

energy contribution and if they are antiparallel (±) and (∓) they cancell each other andgive no energy contribution. Let then

E0(σ) = −J∑(i,j)

σiσj (3.4)

where J is the interactions constant with unit of energy, this is called the nearest-

Neighbour Ising Model. From eq. (3.4) we note that the lowest energy state occurs

when all spins are aligned, meaning the model ”favours” alignment. Meaning it can

demonstrate a spontaneous magnetization i.e. it is a ferromagnetic model.

Lastly from eq. (2.4) we get the energy from particle interaction with the field to

E1(σ) = −H∑i

σi (3.5)

where H is the relation between the direction of field and the magnetic moment.

3.2 Exactly solvable models

Exactly solvable models is often but not necessary models describing completely integrable

systems, informally meaning that quantities in our case the free energy can be expressed

in a integral form. In the field of statistical mechanics the concept of exactly solved

models is more or less synonyms to quantum integrable models where it is possible by

using Yang baxter equation to establish Transfer matrices from which one can express the

Hamiltonian of the system [9].

9

4 The One-dimensional Ising Model

In this chapter we will go through the transfer matrix method of solving the one-

dimensional Ising model. The method of finding the partition function will later be

repurposed when calculating the partition function for the Square-lattice Ising model.

Lastly the mean magnetization can easily and directly be determined from the free energy

and by examining its analyticity we will find that there can not exist a phase transition

for this model.

4.1 The Partition Function

Consider a one-dimensional lattice configuration of N sites, we label each site with a

integer j and we assume periodic boundary conditions such that σ1 = σN+1. This can be

thought of as connecting the end point to form a circle as following: This can be thought

of as connecting the end point to form a circle as following:

1 2 3 N

1

N

2

3

a. b.

Figure 3: Solid lines denotes interaction. (a) One dimensional lattice configuration, (b)

one dimensional lattice configuration with periodic boundary condition

The periodic boundary conditions except from simplifying the solutions of the model,

is the only valid finite version of the system we can consider as closed i.e. no source terms

in the end-points.

Furthermore we assume a nearest-neighbour interaction, we assign a spin variable σj

to each site from eq. (3.4) the interaction energy is then

E0 = −JN∑j

σjσj+1 (4.1)

and we allow the system to be subject to an external field. From eq. (3.4) the field

interaction energy becomes then:

E1 = −HN∑j

σj (4.2)

10

Lastly from eq. (3.2) the energy for each spin-configuration becomes

E(σ) = −JN∑j

σjσj+1 −HN∑j

σj (4.3)

and the partition function ZN follows from eq. (2.2) as

ZN =∑σ

exp[k

N∑j

σjσj+1 + hN∑j

σj

](4.4)

with k = βj and h = βH and where∑

σ is the summation over all possible spin-

configurations.

4.2 The Transfer Matrix V

We want now to rewrite the partition function in a way so we can drop all the summations

from eq. (4.4) and write the parition functions on a form that the free energy can be

calculated. We do this by firstly expanding the sum in the Boltzmann weight as

ZN =∑σ

exp[k(σ1σ2 + σ2σ3 + ...) + h(σ1 + σ2 + ...)

]=

and we split the h-terms

=∑σ

exp[kσ1σ2 + kσ2σ3 + ...+

h

2σ1 +

h

2σ1 +

h

2σ2 +

h

2σ2 + ...)

]=

and express the partition function as a product of Boltzmann weight for just the near-

neighbouring interactions:

=∑σ

exp[kσ1σ2 +

h

2(σ1 + σ2)

]· ... · exp

[kσNσ1 +

h

2(σN + σ1)

](4.5)

In the last factor we used the periodic boundary condition. If we define the function

V (σ, σ′) such as

V (σ, σ′) = exp[kσσ′ +

h

2(σ + σ′)

](4.6)

using V (σ, σ′) we get the partition function to

ZN =∑σ1

∑σ2

...∑σN

V (σ1, σ2)V (σ2, σ3) · ... · V (σN , σ1) (4.7)

where each summation is over the possible spin state of each spin variable [8].

Consider a 2 by 2 matrix V such that all possible spin configurations of V (σ, σ′) is

represented in the elements, such that V is given by

V =

(V (+, +) V (+, −)V (−, +) V (−, −)

)=

(ek+h e−k

e−k ek−h

)(4.8)

11

and if we then return to our partition function. Each factor V (σj, σj+1)V (σj+2, σj+3)

when summed over respective spin variable can be regarded as the matrix product of V .

We consider

ZN =∑σ1

∑σ2

...∑σN

V (σ1, σ2) · ... · V (σN , σ1) =

and from the definition of the matrix product we can regard this as the matrix V N up

the summation of σ1

=∑σ1

V (σ1, σ1)N (4.9)

and from the definition of the trace of matrix, the eq. (4.9) must becomes Tr(V N) giving

us the partition function as:

ZN = Tr(VN) (4.10)

This is a important result, leaving the partition dependent only on the interactions con-

stants.

With V being a real and symmetric matrix it can be diagonalized as

V = PDP−1 (4.11)

where matrix D is a diagonal matrix with the eigenvalues λ1 and λ2 of V in its diagonal

D =

(λ1 0

0 λ2

)(4.12)

and P a matrix containing the eigenvectors x1 and x2 of V

P =(~x1, ~x2) (4.13)

We insert eq. (4.11) into eq. (5.67), we get the power of V to

V N = PDP−1PDP−1 · ... · PDP−1 = PDNP−1 (4.14)

and use the property of the matrix trace:

Tr(PDNP−1) = Tr(DN) (4.15)

Then clearly we have that the partition function can be written as:

ZN = Tr(DN) = λN1 + λ

N2 (4.16)

12

4.3 Free Energy and Phase Transition

The last step is to calculate the the Helmholtz free energy. From eq. (2.3) we have

−(kBT )−1F = ln(ZN) = ln(λN1 + λN2 ) = ln[λN1

(1 +

λN2λN1

)]and we introduce the free energy per lattice site f = F/N . With the free energy F

related to the size of the system we want a quantity in the thermodynamic limit that is

not proportional to N [8]. We have

−(kBT )−1N−1F = ln(λ1) +N−1 ln[1 +

(λ2λ1

)N ](4.17)

we assume λ1 > λ2. In the thermodynamic limit we expect N → ∞ then the last termin the RHS vanish

limN→∞

N−1 ln[1 +

(λ2λ1

)N ]= 0

giving us the free energy per lattice site to:

f = −kBT ln(λ1) (4.18)

The last step is to calculate the eigenvalues which can be obtained with the charac-

teristic equation

0 = det(V − λ) =

∣∣∣∣∣ek+h − λ e−ke−k ek−h − λ∣∣∣∣∣ = (ek+h − λ)(ek−h − λ)− e−2k

the eigenvalues becomes then

λ± =1

2e−2k(e3k+h + e3k−h ± (e2(3k+h) + e2(3k−h) − 2e6k + 4e2k)1/2) (4.19)

using hyperbolic functions gives us the final form of the eigenvalues:

λ± = ek cosh h± (e2k sinh2 h+ e−2k)1/2 (4.20)

For given interactions constants we get the free energy per lattice site to a functions of

the the temperature T and h.

f(h, T ) = −kBT[ek cosh h+ (e2k sinh2 h+ e−2k)1/2

](4.21)

With equation eq. (2.6) we get the magnetization

M(H, T ) =ek sinh βH

(e2k sinh2 βH + e−2k)1/2(4.22)

we have that M(H, T ) is an analytical function for for real H and positive T . Meaning

no phase transitions for positive temperatures.

13

5 The Square-lattice Ising Model

In this chapter we will cover the solutions of the Square-lattice model, the partition func-

tion is determined in terms of the transfer matrices V and W . In the following chapters

will interpret Baxter Rodney’s solutions method of the commuting transfer matrices [8].

With the Star-triangle transformation a commutation relations of the transfer ma-

trices is found. From commutation of the transfer matrices and the operators R, C

and together with Kramers-Wannier duality a eigenvalues relation is determined. Which

lastly is parametrized with elliptic function and the free energy calculated. The existence

of a phase transition and the critical temperature is found with Kramers-Wannier duality

and analysing singularities in the free energy.

5.1 Partition function Square-lattice Ising model

Consider a square-lattice configuration of N sites under the presence of no external field.

We impose periodic boundary conditions such that end-site of each row and column is

connected, the lattice configuration can be thought of as the surface of a toroidal.

a. b.

Figure 4: (a) Square lattice configuration, (b) square lattice configuration with periodic

boundary condition

Furthermore we assume a nearest-neighbour interaction between the sites, where we

allow anisotropy in the interactions i.e. the interaction constant between horizontal ori-

ented sites is J and between vertical oriented sites it is J ′. This results in four interactions

for one site as following:

J

J

J’

J’

σiσj

σk

Figure 5: Nearest-neighbour interaction in one site

14

We formulate the systems energy as a pairwise summation of the interactions terms

for all sites in horizontal and vertical orientation

E(σ) = −J∑(i,j)

σiσj − J ′∑(i,k)

σiσk (5.1)

and the partition function from eq. (2.2).

ZN =∑σ

exp[K∑(i,j)

σiσj + L∑(i,k)

σiσk

](5.2)

with K = Jβ and L = J ′β.

5.2 The Transfer Matrices V and W

The objective is to formulate a transfer matrix description of the Square-lattice model

similar to the one-dimensional case. We do this by rotating our lattices as seen in fig.

(6).

L L LK K K

K L K L K L

Figure 6: Square lattice configuration with rotated orientation and different colors of the

sites to easier distinguish different rows.

Let the total number of rows be m, number if sites in each row n and denote the spin-

configuration of each row φr were then 1 ≤ r ≤ m and φr = {σ1, ..., σn}.We want to describe how rows interact with near-neighbouring rows, ahead when

considering two rows let φ denote the lower row and φ′ denote the upper row. We then

return to fig. (6) and consider the lowest and the middle row. From fig. (7)

L L LK K K L

σj

K

σj+1 σj+2 σj+3 σj σj+1

σ’j σ’j+1 σ’j+2 σ’j

Figure 7: The interaction between two rows is described as a repetition of the right side

interactions

we identify left-side pattern to be a repetition of the right side interaction, mainly that σj

interacts with σ′j and σj+1 interacts with σ′j. Introduce the Boltzmann weight describing

15

the row to row interaction as the function V (φ, φ′) where:

V (φ, φ′) = exp[ n∑j=1

(Kσj+1σ′j + Lσjσ

′j)]

(5.3)

Further we consider the middle and highest row in fig.(6). We have from fig. (8)

K L K L K L K L

σ’j σ’j+1 σ’j+2 σ’j+3

σj σj+1 σj+2σj

σ’j σ’j+1

Figure 8: The interaction between two rows is described as a repetition of the right side

interactions, we note the the interaction share the middle row with fig. (7)

that we make the a similar conclusion and introduce the Boltzmann weight in this case

W (φ, φ′) where:

W (φ, φ′) = exp[ n∑j=1

(Kσjσ′j + Lσjσ

′j+1)

](5.4)

The strength of this formalism is that the partition function becomes a consecutive

product of the functions V (φ, φ′) and W (φ, φ′) as

ZN =∑φ1

∑φ2

...∑φm

V (φ1, φ2)W (φ2, φ3) · ... · V (φm−1, φm)W (φm, φ1) (5.5)

where now sum for over the spin configuration of each row and we used the periodic

boundary conditions in the last weight.

We introduce the matrices V and W with elements Vφ,φ′ and Wφ,φ′ such that each

element is one possible spin configuration of V (φ, φ′) or W (φ, φ′). With 2n possible spin

configuration for a row we get V and W to be 2n by 2n matrices. This is our transfer

matrices for the Square-lattice model, we can write the partition function as

ZN =∑φ1

∑φ2

...∑φm

Vφ1, φ2Wφ2, φ3 · ... ·Wφm, φ1 =

=∑φ1

(VWV · ... ·W )φ1φ1 =∑φ1

((VW )m/2)φ1, φ1 =

= Tr (VW )m/2 (5.6)

where we used the definition of the matrix product to rewrite the first summation and

the definition of the trace of a matrix to rewrite the last summation.

Now the product VW is not symmetric, but in general we have that the trace of a

matrix is the sum of its eigenvalues and from eq (2.2) we get the partition function

ZN = Λm1 + Λ

m2 + ...+ Λ

m2n (5.7)

16

where Λ21, ,Λ22, ... is the eigenvalues of VW .

We want now to calculate the eigenvalues of the matrix VW , to do so we need to

determine a set of properties of the product.

5.3 Star-Triangle Transformation and Commutation Relation

First let V andW instead be functions of the interactions terms as V (K, L) andW (K, L),

this choice will become apparent in next chapters. Furthermore we want to allow a more

general interaction, consider the product

V (K, L)W (K ′, L′) (5.8)

where K, L, K ′ and L′ is now any complex constants and denote the rows as following:

L L LK K K

K’ L’ K’ L’ K’ L’

Figure 9: Square lattice configuration with four interaction constants, with rows labeled

φ, φ′ and φ′′

The elements of the matrix product from eq.s (5.3) and (5.4) is given by

(VW )φ, φ′ =∑φ′′

Vφ, φ′′Wφ′′, φ′ =

=∑φ′′

exp[ n∑j=1

(Kσj+1σ′′j + Lσjσ

′′j )]

exp[ n∑j=1

(K ′σ′′j σ′j + L

′σ′′j σ′j+1)

]=

=∑φ′′

N∏j=1

exp[σ′′j (Kσj+1 + Lσj +K

′σ′j + L′σ′j+1)

](5.9)

Where the spin variable σ′′j = ±1 will only for every factor change the overall sign of theexponents, it is useful to define

(VW )φ, φ′ =N∏j=1

X(σj, σj+1; σ′j, σ

′j+1) (5.10)

with the function

X(σj, σj+1; σ′j, σ

′j+1) =

∑σ′′

exp[(σ′′j (Kσj+1 + Lσj +K

′σ′j + L′σ′j+1)

]= 2 cosh(Kσj+1 + Lσj +K

′σ′j + L′σ′j+1) (5.11)

17

where we expanded the sum and used the definition of hyperbolic function cosh x. More

important we have that X(σj, σj+1; σ′j, σ

′j+1) describes a single star-interaction as seen

in fig. (10).

L K

K’ L’

σj σj+1

σ’j σ’j+1

σ’’j

Figure 10: Graphic representation of the interaction described by X(σj, σj+1; σ′j, σ

′j+1)

The main idea of this solution of the square-lattice Ising model is to establish that V

and W both commute i.e. we ask whether V and W satisfy the equation:

V (K, L)W (K ′, L′) = V (K ′, L′)W (K, L) (5.12)

To anwer this we need to examine a different lattice configuration, consider a three-vertex

interaction as seen in fig. (11).

B1

σi σj

σl

σk

B3

B2

Figure 11: Two dimensional lattice configuration with three-vertx

We want to find a function that similar to X(σj, σj+1; σ′j, σ

′j+1) that describes the inter-

action, introduce the function w such as:

w(σi, σj, σk) =∑σl

exp[σl(B1σi +B2σj +B3σk)

]=

= 2 cosh(B1σi +B2σj +B3σk) (5.13)

We note that w(σi, σj, σk) firstly is unchanged if we negate all the spins. Also for eq.

(5.13) there must exist constants R, C1, C2, C3 such that

w(σi, σj, σk) = R exp[C1σiσj + C2σjσk + C3σkσi

](5.14)

which is describing the system as a interaction in a triangle.

18

This is called the star-triangle transformation. It is graphically represented in fig. (12).

B1

σi σj

σl

σk

B3

B2

σi

σk

σjC1

C2 C3

ST-trans.

Figure 12: Performing ST-transformation on a vertex.

Now if we set eq. (5.13) equal to eq. (5.14) and evaluate for all possible spin configurations

w(+, +, +) = 2 cosh(B1 +B2 +B3) = R exp[C1 + C2 + C3] (5.15)

w(+, +, −) = 2 cosh(B1 +B2 −B3) = R exp[C1 + C2 − C3] (5.16)w(−, +, +) = 2 cosh(−B1 +B2 +B3) = R exp[−C1 + C2 − C3] (5.17)w(+, −, +) = 2 cosh(B1 −B2 +B3) = R exp[−C1 − C2 + C3] (5.18)

this the Star-Triangle relations [8]. They relate the interactions terms of different lattice

structures, but for our case they is key to determine if V and W commute.

We return to the interaction in fig. (10) and we add a particle to particle interaction

M such that we form the triangle which result in a interaction as in fig. (13),

L K

K’ L’

σj σj+1

σ’j σ’j+1

σ’’jM

Figure 13: Square lattice interaction with added interaction, we will use the triangle in

the ST-relations

which can be described by the function w1 where:

w1(σj, σj+1; σ′j, σ

′j+1) =

∑σ′′

exp[Mσjσ

′j + Lσjσ

′′j +Kσj+1σ

′′j + L

′σ′j+1σ′′j +K

′σ′jσ′′j

]=

= eMσjσ′j

∑σ′′

exp[σ′′j (Lσj +Kσj+1 + L

′σ′j+1 +K′σ′j)]

(5.19)

We note that the added interaction M is isolated in the term eMσjσ′j .

19

We want to investigate eq. (5.12), we interchange L and K with L′ and K ′ and lastly

also add the interaction M which result in a interaction as in fig. (14)

L’ K’

K L

σj σj+1

σ’j σ’j+1

σ’’j M

Figure 14: Square lattice interaction with added interaction and interchanged interaction

constants.

which can be described by the function w2 where:

w2(σj, σj+1; σ′j, σ

′j+1) =

∑σ′′j

exp[Mσj+1σ

′j+1 + Lσ

′j+1σ

′′j +Kσ

′j1σ

′′j + L

′σjσ′′j +K

′σj+1σ′′j

]=

= eMσj+1σ′j+1

∑σ′′j

exp[σ′′j (Lσ

′j+1 +Kσj′ + L

′σj +K′σj+1)

](5.20)

Now we ask whether if the exist M such that w1(σj, σj+1; σ′j, σ

′j+1) = w2(σj, σj+1; σ

′j, σ

′j+1)

mainly if we have

L K

K’ L’

σj σj+1

σ’j σ’j+1

M

L’ K’

K L

σj σj+1

σ’j σ’j+1

M=

then we identify the interaction constants in each of the two triangles in fig. (5.3) and

relate them to corresponding interactions constants in the ST-relations. This gives us

that:

C1 = L C2 = K′ C3 = M (5.21)

We can then perform the ST-transformation which will result in:

B1

B3

σj

σ’j

K

L’

σj+1

σ’j+1

L’

K

σj

σ’j

B3

B1

σj+1

σ’j+1

B2 B2=

Figure 15: ST-transformed triangles

20

If we then compapare this to the ST-relations we identify that

B1 = L′ B3 = K (5.22)

We insert (5.21) and (5.22) in the star-triangle relations and get the following:

2 cosh(L′ +B2 +K) = R exp[L+K′ +M ] (5.23)

2 cosh(L′ +B2 −K) = R exp[L−K ′ −M ] (5.24)2 cosh(−L′ +B2 +K) = R exp[−L+K ′ −M ] (5.25)

2 cosh(L′ −B2 +K) = R exp[−L−K ′ +M ] (5.26)

We then add/subtract eq. (5.23) with eq. (5.26) and we get

4 cosh(B2) cosh(K + L′) = 2ReM cosh(L+K ′) (5.27)

4 sinh(B2) sinh(K + L′) = 2ReM sinh(L+K ′) (5.28)

furthermore we add/subtract eq. (5.24) with eq. (5.25) giving us

4 cosh(B2) cosh(K − L′) = 2Re−M cosh(L−K ′) (5.29)4 sinh(B2) sinh(K − L′) = 2Re−M sinh(L−K ′) (5.30)

Lastly we divide eq. (5.27) with eq. (5.28) and eq. (5.30) with eq. (5.29) and multiply

the result as

cosh(K + L′) sinh(K − L′)cosh(K − L′) sinh(K + L′)

=cosh(L+K ′) sinh(L−K ′)sinh(L+K ′) cosh(L−K ′)

(5.31)

this simplifies to the relation

sinh(2K) sinh(2L) = sinh(2K ′) sinh(2L′) (5.32)

which do not depend on M , we can then set M to something more preferable like zero.

Mainly if K, L, K ′ and L′ satisfies eq. (5.32) we can interchange the interaction constants;

giving us that the commutations relation (5.12) is true.

5.4 Inversion of the Transfer Matrices

Another important property is to express the matrix product in a spin variable free form,

this is done by asking for the inverse of the matrix product. To be more specific we ask

for a given K and L what choices of K ′ and L′ leaves the product VW diagonal.

To ensure diagonality we firstly note that for elements not on the diagonal i.e. φ 6= φ′

we have

σj = σ′j, σj+1 6= σ′j+1 or σj 6= σ′j, σj+1 = σ′j+1 (5.33)

21

and then we must require X = 0. And for elements on the diagonal we have φ = φ′ and

φ = −φ′ because negating elements in V and W before the matrix product leaves theproduct unchanged. We have

σj = σ′j, σj+1 = σ

′j+1 or σj 6= σ′j, σj+1 6= σ′j+1 (5.34)

then we require X 6= 0.Using requirement (5.33) with eq. (5.11) we have

cosh(L+K −K ′ + L′) = 0 (5.35)cosh(L−K −K ′ − L′) = 0 (5.36)

where cosh(z) = 0 have the solutions z = 12iπ(2n + 1), we get for n = 0 that the

interactions constants satisfy

L+K −K ′ + L′ = 12iπ

L−K −K ′ − L′ = 12iπ

solving for K ′ and L′ gives

K ′ = L+1

2iπ, L′ = −K (5.37)

where this is the choice of K ′ and L′ that leaves the product diagonal.

Using requirement (5.34) we have the two cases, firstly when the pairs are alike given

by Xlike as

Xlike = Xlike(σj = σ′j; σj+1 = σ

′j+1)

= 2 cosh(σj(L+K

′) + σj+1(K + L′))

=

= cosh(σj(L+K′)) cosh(σj+1(K + L

′)) + sinh(σj(L+K′)) sinh(σj+1(K + L

′)) =

inserting (5.37) we get

= 2 cosh(σj(2L+

1

2iπ))

and dropping the σj because cosh Z is a even function, we get

Xlike = 2i sinh 2L (5.38)

Secondly when the pairs are unlike given by Xunlike

Xunlike = Xunlike(σj 6= σ′j; σj+1 6= σ′j+1)= 2 cosh

(σj(L−K ′) + σj+1(K − L′)

)=

= cosh(σj(L−K ′)) cosh(σj+1(K + L′)) + sinh(σj(L−K ′)) sinh(σj+1(K − L′)) =

22

inserting (5.37) we get

= − sinh(σj1

2iπ) sinh

(σj+12K

)we have sinh Z is an uneven function and breaking out σj and σj+1 gives us the change

in sign, we get

Xunlike = −2iσjσj+1 sinh 2K (5.39)

Using these results the matrix elements becomes of VW becomes

(2i sinh 2L)nδ(σ1, σ′1) ... δ(σn, σ

′n) + (−2i sinh 2K)nδ(σ1, −σ′1) ... δ(σn, −σ′n) (5.40)

where the factors δ(σ1, σ′1) ... δ(σn, σ

′n) ensures that the term vanish if all the pairs is not

alike and the factors δ(σ1, −σ′1) ... δ(σn, −σ′n) ensures that the term vanish if all the pairsis not unlike. We can write this in matrix form

V (K, L)W (L+1

2iπ, −K) = (2i sinh 2L)nI + (−2i sinh 2K)nR (5.41)

where we used eq. (5.37) and I is the 2n by 2n identity matrix and R is the 2n by 2n

matrix with elements

Rφφ′ = δ(σ1, −σ′1) ... δ(σn, −σ′n) (5.42)

Eq. (5.41) is very useful relations and the RHS of eq. (5.41) as desired ensures a inverse

to the matrix product.

5.5 The Negation Operator R

We want to further discuss the matrix R. This is a useful operator which shifts the sign of

the interaction terms, we can see this if we consider V (K, L)R where the matrix elements

is given by:(V (K, L)R

)φ, φ′

=∑φ′′

Vφ, φ′′Rφ′′, φ =

=∑φ′′

exp[ n∑j=1

(Kσj+1σ′′j + Lσjσ

′′j )]δ(σ′′1 , −σ′1) ... δ(σ′′n − σ′n) =

The only non-vanishing terms is the when σ′′j = −σ′j, we get

= exp[ n∑j=1

(−Kσj+1σ′j − Lσjσ′j)]

this gives us that

V (−K, −L) = RV (K, L) = V (K, L)R (5.43)

and similarly can be shown for W (K, L). Also we have the elements of the inverse

operator R−1

Rφ, φ′ = δ(−σ1, σ′1) ... δ(−σn, σ′n) (5.44)

23

clearly the relation

V (K, L) = R−1V (K, L)R (5.45)

holds. Again similarly can be shown for W (K, L). The significance of eq. (5.45) is that

V (K, L) and W (K, L) then both commutes with R.

5.6 The Shift Operator C

We want to develop further relations for how V and W changes under alteration of the

interactions terms. Consider that for V (K, L) we interchange σj, σ′j and K, L.

L K

σj σj+1

σ’j

K L

σj σj+1

σ’j

K L

σ’j σ’j+1

σj

a. b.

Figure 16: Interaction described by V (K, L), (a) interchange K and L, (b) interchange

σj and σ′j. This result in a interaction described by W (K,L)

By interchanging σj, σ′j is the same as switching the indices Vφ′φ i.e. transpose the matrix

V and as we can se in fig. (16) by also switching K and L we result in W (K, L). This

represent:

W (K, L) = V T (L, K) (5.46)

And by performing the same interchange for V (K, L)W (K, L).

L K

K L

σj σj+1

σ’j σ’j+1

a, b.

K L

L K

σ’j σ’j+1

σj σ j+1

Figure 17: Interaction described by V (K, L)W (K, L), (a) interchange K and L, (b)

interchange σj and σ′j. Leaves the interaction unchanged.

From fig. (17) together with previous reasoning we have

V (K, L)W (K, L) =[V (L, K)W (L, K)

]T(5.47)

or to be more clear this transformation leaves the interaction unchanged.

We want to represent these results as a operator by introducing the 2n by 2n matrix

operator C with elements:

Cφ, φ′ = δ(σ1, σ′2) ... δ(σn, σ

′1) (5.48)

24

The operator C shift the spin label one column [8]. We consider V (K, L)C where the

elements is given by:(V (K, L)C

)φ, φ′

=∑φ′′

Vφ, φ′′Cφ′′, φ′ = (5.49)

=∑φ′′

exp[ n∑j=1

(Kσj+1σ′′j + Lσjσ

′′j )]δ(σ′′1 , σ

′2) ... δ(σ

′′n, σ

′1) =

we have that the only non-vanishing terms is when σ′′j = σ′j+1, we get

= exp[ n∑j=1

(Kσj+1σ′j+1 + Lσjσ

′j+1)

](5.50)

This can be graphically represented as seen in fig. (18)

K L K

σ’j-1 σ’j

σj σj+1

K L K

σ’j σ’j+1

σj σj+1

C

Figure 18: Transformation with the operator C, which results in the interaction W (K, L).

meaning shifting the spin variables gives us the relation:

W (K, L) = V (K, L)C (5.51)

Furthermore we also have the inverse operator C−1 with elements given by

C−1φ, φ′ = δ(σ2, σ′1) ... δ(σ1, σ

′n) (5.52)

and if we consider C−1V (K, L)C with elements is given by(C−1V (K, L)C

)φ, φ′

=∑φ′′

∑φ′′′

C−1φ, φ′′Vφ′′, φ′′′Cφ′′′, φ′ =

=∑φ′′

∑φ′′′

δ(σ2, σ′′1) ... δ(σ1, σ

′′n) exp

[ n∑j=1

(Kσ′′j+1σ′′′j + Lσ

′′j σ′′′j )]·

· δ(σ′′′1 , σ′2) ... δ(σ′′′n , σ′1)

we have that the only surviving terms is when σ′′j = σj+1 and σ′′′j = σ

′j+1 we get

= exp[ n∑j=1

(Kσj+2σ′j+1 + Lσj+1σ

′j+1)

]= exp

[ n∑j=1

(Kσj+1σ′j + Lσjσ

′j)]

not surprisingly V (K, L) remains unchanged. In other words we have established

V (K, L) = C−1V (K, L)C (5.53)

25

and similar can be said for W (K, L). We have the V and W also commute with C.

From this result we can simplify the matrix product VW , we have as long as K, L

and K ′ and L′ satisfies relation (5.32)

sinh 2K sinh 2K = sinh 2K ′ sinh 2L

we have then that eq. (5.12):

V (K, L)W (K ′, L′) = V (K ′, L′)W (K, L)

With eq. (5.51) this can be written as

V (K, L)V (K ′, L′) = V (K ′, L′)V (K, L) (5.54)

The same can be said for W (K, L). Eq. (5.54) states that V and W commute with them

self. Furthermore from eq. (5.41) we have then

V (K, L)V (L+1

2iπ, −K)C = (2i sinh 2L)nI + (−2i sinh 2K)nR (5.55)

and again similarly for W (K, L).

5.7 Alternate Form of the Transfer Matrices

At this point it is unnecessary to continue consider the sum over the spin variables in

V (K, L). We are only interested in the overall number of changes in sign and not the

specific sign in each site. Let

r = number of unlike pairs of (σj+1, σ′j) (5.56)

s = number of unlike pairs of (σj, σ′j) (5.57)

Now return to V (K, L) and expand the sum

exp[ n∑j=1

(Kσj+1σ′j +Lσjσ

′j)]

= exp[K(σ2σ

′1 + σ3σ

′2 + ...) +L(σ1σ

′1 + σ2σ2 + ...)

](5.58)

for every unlike pair (σj+1, σ′j) we will have contribution of a (−1) in the parenthesis

canceling out a like pair, with n pair in total and r changes in sign we get (n − 2r)surviving K terms. Similarly for the unlike pairs (σj, σ

′j), given us (n − 2s) surviving L

terms.

We have now the elements for the transfer matrix V can be expressed without the

spin variables as

V (K, L) = exp[(n− 2r)K + (n− 2s)L] = (5.59)

26

in the thermodynamic limit we can assume without loss of generality n to be even. Let

n = 2p, then

= exp[2(p− r)K + 2(p− s)L]

this simplifies the result if we introduce r′ and s′ such that

V (K, L) = exp[±2r′K ± 2s′L] (5.60)

where r′ and s′ are positive integers. The (±1) arises from that exp[2r′K+2s′L] representtwo spin configurations where the other one is obtained by negating all spin variable. As

discussed in Appendix A. both r and s must be even, this gives us that s′ and r′ is both

be either even or odd. Then if we consider

exp[2K]r′exp[2L]s

′

and negating (− exp[2K]

)r′ (− exp[2L])s′will then leave the RHS of eq. (5.60) unchanged. This condition can be expressed as

allowing K = K ± 12πi which will yield exp[2K ± πi] = − exp[2K] and same for exp[2L].

This gives us

V (K ± 12iπ, L± 1

2iπ) = V (K, L) (5.61)

and similarly for W (K, L).

5.8 Functional Relation of the Eigenvalues

We have now all the necessary properties of the transfer matrices to calculate the eigen-

values of the matrix product VW . Firstly we define

k =(

sinh 2K sinh 2L)−1

(5.62)

where k is a real number. We have from Appendix A. that for T = Tc then k = 1. We

regard now the complex variables K and L constrained to a fixed k given by eq. (5.62).

By variation of K and L we can generate transfer matrices V (K, L).

Furthermore we have established the relations

V (K, L) commutes with itself

V (K, L), C commute

V (K, L), R commute

the same can be demonstrated for W (K, L). For commuting matrices there must be a

common set of eigenvectors [10]. Then all matrices V (K, L) and W (K, L) for which K

and L satisfies (5.62) must share at least a common eigenvector x with C and R.

27

We have that x is a function of K and L but for every K and L we can assign a k,

so we regard the eigenvector instead as a function x(k). Further let v(K, L), c, r be the

corresponding eigenvalues of V (K, L), C, R. We have then

V (K, L)x(k) = v(K, L)x(K)

Cx(k) = cx(k)

Rx(k) = rx(k) (5.63)

If we consider Cn which corresponds to shifting a spin label σj with σj+n which due

to the periodic boundary condition is σj = σj+n, we get Cn = I. Also we have R2 which

corresponds to negating a spin label twice which gives us R2 = I. The eigenvalues of Cn

and R2 must therefore satisfy:

cn = r2 = 1 (5.64)

We return the to eq. (5.55), multiplying both side with x(k) then

V (K, L)V (L+1

2iπ, −K)Cx(k) = (2i sinh 2L)nIx(k) + (−2i sinh 2K)nrx(k)

we get then from the commutation relations

v(K, L)v(L+1

2iπ, −K)c = (2i sinh 2L)n + (−2i sinh 2K)nr (5.65)

We recall V (K, L)W (K, L) have the eigenvalues Λ21, Λ22, ...., Λ

22n . Using eq. (5.51), we

write the matrix product as

V (K, L)W (K, L) = V (K, L)V (K, L)C = V 2(K, L)C (5.66)

where

V 2(K, L)Cx(k) = v2(K, L)cx(k) (5.67)

the LHS of eq. (5.67) is our desired eigenvalues

Λ2(K, L) = v2(K, L)c

Λ(K, L) = v(K, L)c1/2 (5.68)

inserting eq. (5.68) to eq. (5.65) gives us the eigenvalues relation:

Λ(K, L)Λ(L+1

2iπ, −K) = (2i sinh 2L)n + (−2i sinh 2K)nr (5.69)

Now with k given by eq.s (5.62) and (5.69) we can not intuitively but through a

straightforward parametrization calculate the eigenvalues of VW .

28

5.9 Parametrization of K and L

Firstly we consider the case k < 1 for reasons that soon will be apparent and return to

eq. (5.60) the elements of V can be written in a exponential form

exp[±2r′K ± 2s′L] = exp[±2K]r′ exp[±2L]s′ (5.70)

For a given k we want to parametrize eq. (5.62) such that exp(±2K) and exp(±2L) issingle-valued meromorphic function i.e. analytic over a given domain with exception for

isolated singularities each of which is a pole. Introduce the parametrization

sinh 2K = x (5.71)

sinh 2L = (kx)−1 (5.72)

which satisfies eq. (5.62). We then solve for exp(2K) and exp(2L)

2K = arcsinh x = ln(x+ (x2 + 1)1/2)⇔⇔ exp(2K) = x+ (x2 + 1)1/2 (5.73)

and

2L = arcsinh (kx)−1 = ln((kx)−1 + ((kx)−1 + 1)1/2)⇔⇔ exp(2L) = (kx)−1[1 + (1 + (kx)2)1/2] (5.74)

but the terms (x2 + 1)1/2 and (1 + (kx)2)1/2 make exp(2K) and exp(2L) not single-

valued and for a complex number z then w =√z is not meromorphic over the complex

plane. There is no parametrization with elementary functions of a general k that makes

(x2 + 1)1/2 and (1 + (kx)2)1/2 perfect squares. However, such parametrization can be

made with elliptic functions [8].

5.10 Parametrization with Elliptic Functions

A elliptic function is a meromorphic function i.e. a functions which is analytic on its

domain except for isolated singularities which are poles. Moreover a elliptic functions

defined for two variables and is periodic in two directions. Compared to a real function

which is defined by the values in a intervall, the elliptic functions is defined by the values

lying in a parallelogram. In our case a rectangle further called a period-rectangle [11].

Firstly we introduce the Jacobi elliptic functions

sn u = k−1/2H(u)/Θ(u)

cn u = (k′/k)1/2H1(u)/Θ(u) (5.75)

dn u = k′1/2Θ1(u)/Θ(u)

29

where Θ(u), Θ1(u), H(u) and H1(u) is the theta functions [8, P. 456]. The Jacobi elliptic

function have two important relations:

cn2 u = 1− sn2 u (5.76)dn2 u = 1− k2 sn u (5.77)

From which if set the parametrization to

x = −i sn(iu) (5.78)

we can resolve the square roots such that the result is meromorphic. With eq.s (5.78),

(5.76) and (5.77) we get exp(±2K) and exp(±2L)

exp(±2K) = ∓i sn(iu) + (1− sn2(iu))1/2 = ∓ sn(iu) + cn(iu)exp(±2L) = (∓ik)−1(1 + (1− k2 sn( iu))1/2) = ik−1(dn(iu)± 1)/ sn(iu) (5.79)

Using eqs. (5.75) we can express exp(±2K) and exp(±2L) in terms H, H1, Θ and Θ1which are function entire functions i.e. analytical everywhere [8]. Resulting in

exp(±2K) = [k′1/2H1(iu)∓ iH(iu)]/[k1/2Θ(iu)] (5.80)exp(±2L) = i[k′1/2Θ1(iu)±Θ(iu)]/[k1/2H(iu)] (5.81)

The fraction together with eq. (B.7) from Appendix B. gives us that exp(±2K) andexp(±2L) is meromorphic functions

The reason of only considering k < 1 is because as from Appendix B. we find the

elliptic functions only to be defined for 0 < k < 1. We will later return to case when

k = 1 and k > 1.

5.11 The form of Λ(u)

Next step is to determine the form of eigenvalues in terms of the elliptic functions. We

have from earlier the form of the matrix element of V given by

V (K, L) = exp[±2K]r′ exp[±2L]s′ =

using eq. (5.80), (5.81) and r′ = p− r, s′ = p− s we get

=...

[k1/2Θ(iu)]p−r[k1/2H(u)]p−s= (5.82)

currently we are only interested in the structure of the poles, let the numerator be . . .

which is just a entire function

=· · ·

[Θ(iu)H(iu)]pΘ−r(iu)H−s(iu)

30

where we let the k factors be a part of . . . , final step gives us

V (K, L) =· · ·

[h(iu)]p(5.83)

where h(u) = H(u)Θ(u) and now using the relations (5.63) and eq. (5.68) we get:

Λ(u) =· · ·

[h(iu)]p(5.84)

One advantage of the elliptic functions is their periodicity under their half amplitudes I

and I ′. We want to determine how eq.s (5.79) change if we set u → u + 2I ′. Using theperidocicity relations

sn(u+ 2iI ′) = sn(u), cn(u+ 2iI ′) = − cn(u), dn(u+ 2iI ′) = − dn(u) (5.85)

We get the RHS. of eq.s (5.79) to

∓ sn(iu+ 2iI ′) + cn(iu+ 2iI ′) = ∓ sn(iu)− cn(iu)ik−1(dn(iu+ 2iI ′)± 1)/ sn(iu+ 2iI ′) = ik−1(dn(iu)± 1)/ sn(iu)

Eq.s (5.79) is satisfied if K → −K ± 12iπ and L→ −L± 1

2iπ, we have

exp(∓2K + iπ) = ∓ sn(iu)− cn(iu)exp(∓2L+ iπ) = ik−1(− dn(iu)± 1)/ sn(iu)

meaning:

u→ u+ 2I ′ ⇔

K → −K ± 12iπL→ −L± 12iπ

(5.86)

We have then using eq. (5.61) and then (5.43)

V 2(−K ± 12iπ, −L± 1

2iπ) = V 2(−K, −L) = R2V 2(K, L)

and taking the square-root and multiplying both sides with x(k) give us:

Λ(−K ± 12iπ, −L± 1

2iπ) = rΛ(K, L) (5.87)

or expressed as a function of u

Λ(u+ 2I ′) = rΛ(u) (5.88)

where the eigenvalue r is then r = ±1. Furthermore we want to determine how eq.s(5.79) change if we set u→ u− 2iI. Using the relations

sn(u+ 2I) = sn(u), cn(u+ 2I) = − cn(u), dn(u+ 2I) = dn(u) (5.89)

31

We get the RHS. of eq.s (5.79) to

∓ sn(iu+ 2I ′) + cn(iu+ 2I ′) = ± sn(iu)− cn(iu)ik−1(dn(iu+ 2I ′)± 1)/ sn(iu+ 2I ′) = −ik−1(dn(iu)± 1)/ sn(iu)

Eq.s (5.79) is satisfied if K → K ± 12iπ and L→ L± 1

2iπ, we have

exp(±2K + iπ) = ± sn(iu)− cn(iu))exp(±2L+ iπ) = −ik−1(dn(iu)± 1)/ sn(iu)

meaning:

u→ u− 2iI ⇔

K → K ± 12iπL→ L± 12iπ

(5.90)

by a similar discussion as earlier, we get the Λ(u) satisfying

Λ(u− 2iI) = Λ(u) (5.91)

From eq.s (B.1) we have eq. (5.88) and (5.91) ensures that Λ(u) is a doubly periodic

function see Appendix B. together with theorem (C.3) the function Λ(u) must be on the

form

Λ(u) = ρe−λun∏j=1

[H(u − uj)]/H(u − vj)] (5.92)

from eq. (5.84) we know the structure of the poles. We have

2p∏j=1

1/H(u − vj) = [h(iu)]−p (5.93)

letting the minus sign be a part of λ gives us.

Λ(u) = ρeλu[h(iu)]−p2p∏j=1

H(iu− iuj) (5.94)

where u1, ..., u2p are the zeros of Λ(u) within a period rectangle, ρ and λ ensure periodicity.

5.12 Zeros of Λ(u)

The last step is to determine the zeros of Λ(u). We return to eq. (5.69):

Λ(K, L)Λ(L+1

2iπ, −K) = (2i sinh 2L)n + (−2i sinh 2K)nr

We want to express the relation as a function of u, to do this we ask how eq.s (5.79)

change for u→ u+ I. Using the relations

sn(u+ iI ′) = (k sn(u))−1

cn(u+ iI ′) = −i dn(u)/(k sn(u))dn(u+ iI ′) = −i cn(u)/ sn(u) (5.95)

32

we get the RHS of eq.s (5.79) to

∓ sn(iu+ iI ′) + cn(iu+ iI ′) = ik−1s(− dn(iu)∓ 1

)/ sn(iu)

ik−1(

dn(iu+ iI ′ ± 1)/ sn(iu+ iI ′))

= ±i sn(iu) + cn(iu) (5.96)

and this is satisfied if K → L+ 12iπ and L→ −K such that

exp(±L± iπ) = ik−1s(− dn(iu)∓ 1

)/ sn(iu)

exp(∓K) = ±i sn(iu) + cn(iu)

meaning that

u→ u+ I ′ ⇔

K → L+ 12iπL→ −K (5.97)the RHS of (5.69) can be written as Λ(u)Λ(u + I ′). Moreover we have from eq.s (5.76),

(5.77) and (5.78) we get the relation between K, L and u from:

sinh 2K = −i sn(iu) (5.98)

sinh 2L =i

k sn(iu)(5.99)

Inserting this in the eigenvalue relation we get:

Λ(u)Λ(u+ I ′) =(− 2k sn(iu)

)n+ (−2 sn(iu))nr (5.100)

We now have the relation on a u dependent form, we can also rewrite Λ(u)Λ(u+ I ′).

Using n = 2p and eq. (5.94) the LHS of eq. (5.100) becomes

Λ(u)Λ(u+ I ′) = ρeλu[h(iu)]−p2p∏j=1

H(iu− iuj) · ρeλ(u+I′)[h(i(u+ I ′))]−p

2p∏j=1

H(iu− iuj + iI ′) =

= ρ2eλ(2u+I′)[h(iu)]−2p

2p∏j=1

H(iu− iuj)H(iu− iuj + iI ′) (5.101)

where we used from Appendix B. eq. (B.10) which gives us h(i(u + I ′)) = h(iu). Fur-

thermore the RHS of eq.(5.100) combined with eq (5.75) gives us:(− 2k sn(iu)

)n+ (−2 sn(iu))nr = (2/k)p

[(Θ(iu)H(iu)

)2p+

(H(iu)rΘ(iu)

)2p]

=

= (2/k)p(

Θ4p(iu) +H4p(iu)r)

(h(iu))−2p (5.102)

These results give us the relation

ρ2eλ(2u+I′)

2p∏j=1

H(iu− iuj)H(iu− iuj + iI ′) = (2/k)p(

Θ4p(iu) +H4p(iu)r)

(5.103)

this gives us an overall restriction to the zeros of Λ(u). We ask for what uj and u the

LHS and RHS vanish.

33

5.13 Zeros of the RHS

Firstly we investigate when the RHS of eq. (5.103)

(2/k)p(

Θ4p(iu) +H4p(iu)r)

vanish. This occur when (− 2k sn(iu)

)n+ (−2 sn(iu))nr = 0

this can be simplified by

1 + (k sn2(iu))nr = 0

r + (k sn2(iu))nr2 = 0

where we used r2 = 1 and n = 2p giving us

k sn2(iu))2p + r = 0 (5.104)

If we consider sn(u) we have from the relations (5.85) and (5.89) that it is doubly

periodic with 2I and 2iI ′. This gives us that sn(u) have a period rectangle with width

2I and hieght 2iI ′. Moreover from eq. (5.75) sn(u) have a pole when Θ(u) = 0, from

Appendix B. eq. (B.7) this occur when u = 2mI + i(2n − 1)I ′ where n and m is anyintegers. We return to (k sn2(iu))2p + r there is then only one pole of order 4p per period

rectangle. From theorem (C.2) we have that (k sn2(iu))2p + r have one zero per period

rectangle. What is then the structure of u?

It is to our advantages to set

u = −12I ′ − iφ (5.105)

we get then that the zeros occur when:

(k sn2(φ− 12iI ′))n + r = 0 (5.106)

From this we can use the elliptic amplitude function

Am(φ) = − ln[ik1/2 sn(φ− 1

2iI ′)]

(5.107)

with the above mentioned gives us that the zeros occur instead when:

exp[4ipAm(φ)] + r = 0 (5.108)

Eq. (5.108) have the solutions

Am(φ) =

12pπn+ 14pπ r = 112pπn r = −1

(5.109)

34

where n is an integer. We define

θj =

12pπ(j − 12) r = 112pπj r = −1

(5.110)

where j = 1, ..., 2p we have that eq. (5.108) is satisfied by φ = φj if

Am(φ) = θj −1

2π (5.111)

where we add −12π for reason that soon will be clear.

Now the advantages of expressing the solutions in terms of the amplitudes function

comes from its behaviour between −I and I. We have from the taylor expansion of Am(φ)[8]:

Am(φ) =π

2Iφ + 2

∞∑m=1

qm/2

m(1 + qm)sin(mπIφ)

(5.112)

Where q is from Appendix B. Most important must Am(φ) increases monotonically from

−12π to 1

2π as φ increases from −I to I and we have: π4p ≤ θj ≤ π(1− 14p) r = 1π

4p≤ θj ≤ π r = −1

(5.113)

Giving us 0 < θj ≤ π, with the shift −12π to θj we ensures a distinct and real solutionsfor every j with −I < φ ≤ I.

Lastly from eq. (5.95) we have by setting u→ u+ I ′

(k sn2(iu+ iI ′))2p + r = 0

1/(k sn2(iu))2p + r = 0 (5.114)

so from eq. (5.114) if u is a solutions so must then u+ I ′ also be a solution, we get

u = ∓12I ′ − iφj (5.115)

for j = 1, ... , 2p.

5.14 Zeros of the LHS

Lastly we investigate when the LHS of eq. (5.103)

ρ2eλ(2u+I′)

2p∏j=1

H(iu− iuj)H(iu− iuj + iI ′)

vanish. From eq. (5.115) we have the zeros of H(iu− iuj) when uj satisfy

uj =1

2γjI′ − iφj

35

where γj = ±1 and j = 1, ..., 2p. However H(iu − iuj + iI ′) guarantees a solution whenuj − I ′ giving us the general solutions to

uj = −1

2γjI′ − iφj (5.116)

but not all solutions of eq. (5.116) are allowed.

From the doubly periodicity of eq. (5.94) and restriction of the zeros in eq. (C.2)

which can be written as

u1 + ...+ u2p = (p+ 2l′)I ′ + i

[12

(1− r) + 2l]

(5.117)

where l and l′ are integers [8]. Then to ensure a double periodicity λ from eq. (C.3) is

determined to:

λ = −πγju/4I (5.118)

Inserting to eq. (5.94) and we get

Λ(u) = ρ[h(iu)]−p2p∏j=1

e−πγju/4IH(iu− φj +1

2iγjI

′) (5.119)

and if we square both sides

Λ2(u) = ρ2[h(iu)]−2p2p∏j=1

e−πγju/2IH2(iu− φj +1

2iγjI

′)

this is the eigenvalues of VW . There exist still room for major simplifications.

From the relations (B.10) we have

iq−1/2Θ(iu− φj −1

2iγjI

′)/H(iu− φj +1

2iγjI

′) = exp(−πγju/2I)

resulting in

Λ2(u) = iq−1/22p∏j=1

Θ(iu− φj − 12iγjI′)H(iu− φj + 12iγjI

′)

H(iu)Θ(iu)(5.120)

introduce the constant ρ′ = iq−1/2 and with eq (5.75) we have

k1/2 sn(iu− φj +1

2iγjI

′)Θ(iu− φj +1

2iγjI

′) = H(iu− φj +1

2iγjI

′) (5.121)

we can write the eigenvalues in a γj free form

Λ2(u) = ρ′2p∏j=1

Θ(iu− φj + 12iγjI′)Θ(iu− φj − 12iγjI

′)

H(iu)Θ(iu)k1/2 sn(iu− φj +

1

2iγjI

′) (5.122)

36

because the numerator Θ(iu − φj + 12iγjI′)Θ(iu − φj − 12iγjI

′) does not depend on the

choice of γj. Introduce D such that

Λ2(u) = D

2p∏j=1

k1/2 sn(iu− φj +1

2iγjI

′) (5.123)

where

D = ρ′2p∏j=1

Θ(iu− φj + 12iI′)Θ(iu− φj − 12iI

′)

H(iu)Θ(iu)(5.124)

This can be simplified by noting D is a doubly periodic function of iu. With

poles of order 2p when iu = 0, I ′

and

zeros of order 4p when iu = φj ± iI ′

As discussed in previous section we have that another function satisfying these require-

ments is:(k sn2 iu)2p + r

(k1/2 sn iu)2p(5.125)

Divide D with eq. (5.125) to form the expression

ρ′2p∏j=1

Θ(iu− φj + 12iI′)Θ(iu− φj − 12iI

′)

H(iu)Θ(iu)× (k

1/2 sn iu)2p

(k sn2 iu)2p + r(5.126)

due to that they share the same zeros and poles this expression will be a entire peridodic

function. From theorem C.1 this function must be a constant, meaning D must satisfy

D ∝ (k sn2 iu)2p + r

(k1/2 sn iu)2p(5.127)

We have then

Λ2(u) = A(k sn2 iu)2p + r

(k1/2 sn iu)2p

2p∏j=1

k1/2 sn(iu− φj +1

2iγjI

′) (5.128)

where A is a constant. From eq. 5.114 if we set u→ u+ I ′ we have

Λ2(u+ I ′) = A(1 + k sn2 iu)2pr

(k1/2 sn iu)2p

2p∏j=1

k−1/2[sn(iu− φj +1

2iγjI

′)]−1 (5.129)

Giving us:

Λ2(u)Λ2(u+ I ′) = A2[ r

(k2 sn iu)2p+ (k sn2 iu)2p + 2

](5.130)

Then squaring both sides of eq. (5.100) gives us

Λ2(u)Λ2(u+ I ′) =(4k

)2pr[ r

(k2 sn iu)2p+ (k sn2 iu)2p + 2

](5.131)

37

we compare eq.s (5.130) and (5.131) and identify A as:

A =√r( 2k1/2

)2p(5.132)

Inserting this back to eq. (5.128)

Λ2(u) = τ[( 2k sn iu

)2p+ (2 sn iu)2pr

]×

2p∏j=1

k1/2 sn(iu− φj +1

2iγjI

′) (5.133)

where τ is:

τ =

+1 r = 1−i r = −1 (5.134)We have now fully determined the eigenvalues of the matrix product for k < 1, this also

concludes our need of elliptic functions.

5.15 General Expression for the Eigenvalues

We want now to write the eigenvalues in a more simpler and preferably a form free of

elliptic functions. From eq.s (5.107) and (5.108) we get

k1/2 sn(φj −1

2iI ′) = − exp(iθj) (5.135)

and from eq.s (5.76) and (5.77)

(cn2 u− 1)(dn2 u− 1) = k2 sn4 ucn2 u sn2 u = k2 sn4 u− sn2 u (k2 + 1) + 1

= sn2 u[k2 sn2 u+

1

sn2 u− (k2 + 1)

](5.136)

using eq. (5.135), this can be written as

cn2(φj −1

2iI ′) dn2(φj −

1

2iI ′) = k−1 exp(iθj)

2[k exp(2iθj) + k exp(−2iθj)− (k2 + 1)

](5.137)

and simplifies to:

cn(φj −1

2iI ′) dn(φj −

1

2iI ′) = −ik1/2 exp(iθj)cj (5.138)

where

cj = ±k−1(1 + k2 − 2k cos(2θj))1/2 (5.139)

We have for sn u the addition relation [8, P. 463].

sn(u − v) = sn u cn v dn v − cn u dn u sn v1− k2 sn2 u sn2 v

(5.140)

38

and if the return to k1/2 sn(iu − φj + 12iγjI′) from eq. (5.133), we set u = iu and

v = φj − 12iI′ and using the previous eq.s (5.135) and (5.138) we get:

k1/2 sn(iu− φj +1

2iγjI

′) =cn iu dn iu − ikcj sn iu

exp (−iθj)− k exp(iθj) sn2(iu)(5.141)

We return now to our initial parametrization from eq.s (5.79)

exp(±2K) = ∓ sn(iu) + cn(iu) (5.142)exp(±2L) = ik−1(dn(iu)± 1)/ sn(iu) (5.143)

we can then easily find from the definitions of hyperbolic functions that:

sinh 2K = −i sn iucosh 2K = cn iu

sinh 2L = i/(k sn iu )

cosh 2L = i dn iu /(k sn iu) (5.144)

Inserting eq.s (5.144) to eq. (5.141) and dropping the the dependence on u we introduce

uj such that

uj =cosh 2K cosh 2L+ cj

exp(iθj) sinh 2K + exp(−iθj) sinh 2L(5.145)

and inserting eq. (5.144) to:

τ[(− 2k sn(iu)

)2p+ (−2 sn(iu))2pr

]= τ(−4)[(sinh 2L)2p + r(sinh 2K)2p] (5.146)

Lastly we note from eq. (5.95) that we have the uj = u−1j . Inserting these results into

eq. (5.133) gives us

Λ2 = τ(−4)p[(sinh 2L)2p + r(sinh 2K)2p]2p∏j=1

(uj)γj (5.147)

We have successfully found the eigenvalues of the of the matrix product.

5.16 Analyticity and validity for k = 1 and k > 1

As previous mentioned when finding the eigenvalues we restricted our solution to k < 1

due to our elliptic parametrization only being defined for k < 1. What remains now after

our parametrization extend our eigenvalues to k = 1 and k > 1, we investigate whether

this is possible for eq. (5.147).

The only term that depends on k is cj in uj from eq. (5.145). We have that (uj)γj in

eq. (5.147) with the numerator

cosh 2K cosh 2L+ cj

39

if γj = −1 andcosh 2K cosh 2L + cj = 0

our eigenvalues become unanalytical. We want to require cj to be positive to avoid this.

For k < 1 we have cj to be positive for every θj which as found lies in the interval

0 < θj ≤ π.For k = 1, we have for 0 < θj < π then cj remains positive. If θj = π we have shown

earlier r = −1 and j = 2p, this will yield c2p = 0For k > 1, we have the two cases when 0 < θj < π again cj is positive, but for r = −1

and j = 2p we have θ2p = π giving us

c2p = (1− k)/k (5.148)

which is negative.

To keep Λ2 analytical we set

cj = k−1(1 + k2 − 2k cos(2θj))1/2 (5.149)

except when r = −1, j = 2p and k > 1. Then we set

cj = −k−1(1 + k2 − 2k cos(2θj))1/2 (5.150)

5.17 Maximum Eigenvalue and the Free Energy

With our partition function we can finally calculate the free energy. If we keep the number

of particles in a row n fixed and let number of rows m in thermodynamic limit satisfy

1� m. From eq. (5.147) we have our partition function

ZN = Λm1 + Λ

m2 + ...+ Λ

m2n

where Λ21, ,Λ22, ... is the eigenvalues of VW , then in the thermodynamic limit the largest

eigenvalue Λ2max will grow making the remaining eigenvalues negligible. We have

ZN ≈ (Λmax)m (5.151)

and also together with eq. (2.3) we get the free energy to

F = −kBT ln(Λmax)m (5.152)

but we are interested in the free energy per lattice site f . The total number of particles

is given by rows times columns i.e. N = m× 2p, we get:

f = − kbT2mp

ln(Λmax)m = −kbT

2pln Λmax (5.153)

We determined the form of the eigenvalues, now we need to maximize Λ2 from eq. (5.147).

With K and L to be given the only parts of Λ2 we can vary is choice of r = ±1 and γj.

40

We ask for what r and γj eq. (5.147) attains its maximum. From eq.s (5.62), (5.139)

and (5.145) we get for real K and L

uju∗j =

cosh 2K cosh 2L+ cjexp(iθj) sinh 2K + exp(−iθj) sinh 2L

(5.154)

and for positive K and L

0 ≤ cj ≤ cosh 2L cosh 2L (5.155)

from this we can establish |uj| ≥ 1 then eq. (5.147) is maximized when γ1 = ... = γ2p =+1. With this requirement we can simplify eq. (5.147).

We have then for r = 1 and eq.(5.147) that θj =π2p

(j − 1

2

)2p∏j=1

exp[iπ

2p

(j − 1

2

) ]sinh 2K + exp

[− i π

2p

(j − 1

2

)]sinh 2L =

= −i(−1)p[(sinh 2L)2p − (sinh 2K)2p

]−1(5.156)

and for r = −1 and eq. (5.147) that θj = π2pj

2p∏j=1

exp[iπ

2pj]

sinh 2K + exp[− i π

2pj]

sinh 2L =

= (−1)p[(sinh 2L)2p + (sinh 2K)2 p

]−1(5.157)

Inserting this to (5.147) gives us

Λ2max =

2p∏j=1

2(cosh 2L cosh 2L+ cj) (5.158)

where this is the form of the maximum eigenvalues. We have to consider the values of r

before we calculate the free energy.

We define

F (θ) = ln 2[cosh 2K cosh 2L + k−1(1 + k2 − 2K cos 2θ)1/2] (5.159)

From Appendix C. we have Perron–Frobenius theorem which gives us that for a real square

matrix with positive entries we can choose the corresponding eigenvector to have strictly

positive components. This can only happen when r = 1, then inserting θ = π2p

(j − 1

2

)we

get

Λ2max =

2p∏j=1

eF [π2p

(j− 1

2

)] (5.160)

solving for ln Λmax give us:

ln Λmax =

2p∑j=1

F[ π

2p

(j − 1

2

)](5.161)

41

Lastly inserting to eq. (5.153). For large p the sum can be replaced by the integral

with the sub-interval length π2p

du to that 0 < θj < π giving us

−f/kBT = (2π)−1∫ π

0

dθF (θ) (5.162)

This the complete expression of the free energy.

5.18 The critical temperature and phase transition

We have from the Kramers-Wannier duality the existence of a phase transition in the

square-lattice model. We want to establish if this phase transition have only one corre-

sponding critical temperature, we return to eq. (5.62) and use K = J/kBT , L = J′/kBT .

We get

k =(

sinh 2J/kBT sinh 2J′/kBT

)−1(5.163)

The critical temperature being a discontinuity in the magnetization M is can also be

regarded as singularities in f . We ask now when is f singular?

We can split up eq. (5.159) in two parts as:

F (θ) = ln 2[cosh 2K cosh 2L + k−1(1 + k2 − 2K cos(θ))1/2]

= ln(

2 cosh 2K cosh 2L)

+ ln(

1 +k−1(1 + k2 − 2K cos 2θ)1/2

cosh 2K cosh 2L

)Due to that we only searching for singularities we expand such ln(1+x) = x− 1

2x2+ 1

3x3+...

and get

F (θ) ≈ ln(

2 cosh 2K cosh 2L)

+k−1(1 + k2 − 2K cos 2θ)1/2

cosh 2K cosh 2L(5.164)

inserting to free energy per lattice eq. (5.162) we have

−f/kBT = (2π)−1∫ π

0

dθ[

ln(

2 cosh 2K cosh 2L)

+k−1(1 + k2 − 2K cos 2θ)1/2

cosh 2K cosh 2L

](5.165)

where we can drop the analytical part

(2π)−1∫ π

0

ln(

2 cosh 2K cosh 2L)

The singularities must then occur in fs where

−fs/kBT = (2π)−1∫ π

0

dθk−1(1 + k2 − 2K cos 2θ)1/2

cosh 2K cosh 2L(5.166)

This can be through rewrites, expansions and dropping analytical parts be written as [8,

P. 121]:

−fs/kBT =(1 + k)(1− k)2

2πk cosh 2K cosh 2Lln∣∣∣1 + k1− k

∣∣∣ (5.167)which gives us that f is singular at k = 1 giving us one critical temperature at

sinh(2J/kBTc) sinh(2J′/kBTc) = 1 (5.168)

42

6 Discussion

This thesis describes the general Ising model and relates it to ferromagnetism. It covers

through the transfer matrix method the solution of the One-dimensional Ising model by

finding the Helmholtz free energy as an analytical function. From the free energy it is

concluded that the One-dimensional Ising model demonstrates no phase transition, due

that the magnetization being a analytical function.

The Square-lattice Ising model is solved with the commuting transfer matrices method.

The solutions is found by establishing with help from the Star-Triangle transformation

and the inverse of the matrix product that the transfer matrices commute with them self,

the negation operator R and the spin variable shift operator C. Together with Kramer-

Wannier duality a eigenvalue relation can be found, which can be parametrized through

elliptic function and the free energy exactly calculated. From the free energy the location

of only one critical temperature at k = 1 is determined.

The significance of an exact solution of the Square-lattice model is that it offers a

understanding of ferromagnetism and a simple analytical description of the phenomena

phase transitions. The Star-triangle transformation and the resulting relations have be-

come vital part of solving other models notably the Chiral Potts models which allows spin

variables to take on higher spin values [12].

There exist different programmes for acquiring the the solutions of the Square-lattice

Ising model, firstly is Onsager’s original solution and also in addition Bruria Kaufman

simplification of Onsager’s work which focus on on spinorial representation of the rotation

group [13]. The cases of the Ising model covered in the thesis concerns solids i.e. fixed

particles, the Ising model can be extended to describe fluids or more specific a lattice

gas. The lattice gas assumes particle to only exists on lattice sites but allows sites to

be unoccupied and particles to change occupations, this is a simple model of density

fluctuations and liquid-gas transformations. A similar model to the Square-lattice Ising

models is The spherical model which is a model on a lattice with an external field, in this

case the spin variable is allowed to take any real value constrained to the average value

of the square of any spin is one [14].

A way to improve the thesis is to increase the understanding on how the Ising model

connects to more general aspects of statistical mechanics. A step to do this is to study

the correlation between the spin variables and how their correlation length connects to

the Scaling hypothesis.

The Square-lattice model with an external magnetic field and the three-dimensional

model remains unsolved. There is hope resolving the three-dimensional Ising model with

Conformal field theory which offers descriptions of the Ising model near the vicinity of

critical points [15]. An method of solving conformal field theory which is a today ongoing

field of research is to solve the three-dimensional Ising model with Conformal bootstrap

43

[16].

To conclude one can obtain an analytical free energy for the One-dimensional nearest-

neighbour Ising model with an external field and the Square-lattice nearest-neighbour

Ising model. For the one-dimensional Ising Model we have no phase transitions, for the

Square Lattice Ising Model we have one phase transition and critical temperature located

at eq. 5.168.

44

References

[1] S.G. Brush. History of the lentz-ising model. Review of Modern Physics, 39(4):883–893, 1967.

[2] K. Huang. Statstical Mechanics. John Wiley and sons, 2nd ed. edition, 1987.

[3] R. Peierls. On ising’s model of ferromagnetism. Mathematical Proceedings of theCambridge Philosophical Society, 32(3):477–481, 1936.

[4] G. Gallavotti. Statstical Mechanics. Texts and Monographs in Physics, 1999.

[5] F. Mandl. Statstical Physics. John Wiley and sons, 2nd ed. edition, 1993.

[6] E.M. Lifshitz L.D. Landau. Statstical Physics. Pergamon Press, 2nd ed. edition,1969.

[7] T. Miyazaki and H. Jin. The Physics of Ferromagnetism, volume 158. Springer,2012.

[8] R.J. Baxter. Exactly Solved Models in Statistical Mechanics. Academic Press, 3rdprint edition, 1989.

[9] G. Mussardo. Statistical Field Theory: An Introduction to Exactly Solved Models inStatistical Physics. OUP Oxford, 2009.

[10] K. Hoffman. Linear Algebra. Prentice-Hall, 2nd ed. edition, 1971.

[11] N.I. Akhiezer and H.H. McFaden. Elements of the Theory of Elliptic Functions.Translations of Mathematical Monographs. American Mathematical Soc., 1970.

[12] H. Au-Yang and J.H.H Perk. Onsager’s star-triangle equation: Master key to inte-grability. In Integrable Systems in Quantum field theory and statistical mechanics.Academic Press, 1989.

[13] R. J. Baxter. Onsager and kaufman’s calculation of the spontaneous magnetizationof the ising model. Journal of Statistical Physics, 145(3):518–548, 2011.

[14] T. H. Berlin and M. Kac. The spherical model of a ferromagnet. Phys. Rev., 86:821–835, Jun 1952.

[15] P. Ginsparg. Applied conformal field theory. 1988, arXiv:hep-th/9108028.

[16] D. Poland et al. S. El-Showk, M.F. Paulos. Solving the 3d ising model with theconformal bootstrap. 1988, arXiv:hep-th/1203.6064v3.

[17] H. A. Kramers and G. H. Wannier. Statistics of the two-dimensional ferromagnet.part i. Phys. Rev., 60:252–262, Aug 1941.

[18] H. A. Kramers and G. H. Wannier. Statistics of the two-dimensional ferromagnet.part ii. Phys. Rev., 60:263–276, Aug 1941.

45

Appendix A

This appedix cover an unformal discussion over how the Karmers-Wanniers Duality can

be used to determine the location of the critical temprature. The relations is the fundation

of the paramatrization from which in this report the Square-lattice Ising model is solved.

Kramers–Wannier Duality

Consider a the partition function for a square lattice

ZN(K, L) =∑{σ}

exp[K∑(i,j)

σjσj + L∑(i,k)

σiσk

](A.1)

Let r be number of unlike pairs of (σi, σj) and s be number of unlike pairs of (σi, σk).

The Boltzmann weight becomes :

exp[K(N − 2r) + L(N − 2s)

](A.2)

We are now going to use a dual lattice. Which means firstly we have our square lattice

Γ, we are now going to add a dual lattice ΓD such that it is a square lattice with sites in

the center of the faces of Γ .

a. b.

Figure 19: (1) square lattice Γ, (2) square lattice Γ in solid line, and dual lattice ΓD in

dashed lines

Instead of regarding the spin labels as sites in Γ we can now think of them as faces in

ΓD. Let now the boundary separating neighbouring faces with different sign be labeled

by a solid line and neighbouring faces with the same sign be label with a dashed line.

+––+

–––+

––––

+++–

++

++++

+++

––

–+++

++++

–

++–

–++

–++–

––+–

+++–

++

++++

+

++–

+––

––++

+++

Figure 20: Example of dual lattice configuration, where boundary between faces indicate

if there is a sign change, note periodic boundry conditions is used at the end points.

46

We have then r horizontal lines and s vertical lines on ΓD. Lets consider the sites on

Γ, there must be an even number of dashed and solid lines into the vertex of each site.

This gives us that must be labeled to link these together and form polygons ex. see fig.

(20)

These polygons divide the system in a spin up and spin down domain. We get the

partition function to:

ZN(K, L) = 2∑P⊂ΓD

exp[K(N − 2r) + L(N − 2s)

](A.3)

where we now sum over all possible polygon configurations on ΓD, the factor two comes

from every set of polygons represent two possible spin configurations one is obtained from

the other by negating all the spins.

We evaluate in the low temperature limit. For T � 1 let K = K∗ and L = L∗, we get

ZN(K∗, L∗) = 2

∑P⊂ΓD

exp[K∗(N − 2r) + L∗(N − 2s)

]= (A.4)

= 2 exp[N(K∗ + L∗)]∑P⊂ΓD

exp[−2K]r exp[−2L]s (A.5)

Further by comparing the low and high temperature expansion one can find the they are

related with the transformations:

tanh K = e−2L∗

(A.6)

tanh L = e−2K∗

(A.7)

which guarantees a singularity i.e. a phase transition Kramers–Wannier duality [17, 18].

We write eq. (A.6) as

sinh 2K

(cosh 2K + 1)=

1− e−4L∗

2 sinh L∗

sinh 2K sinh 2L∗ =1

2

(cosh 2K + 1

)(1− e−4L∗

)(A.8)

and by doing the similar for eq. (A.7), we have in the low temperature expansion that

L∗ and K∗ is large and in the high temperature expansion that L and K is small. By

using this we get

sinh 2K sinh 2L∗ = 1 (A.9)

sinh 2K∗ sinh 2L = 1 (A.10)

if the critical point is located at K = Kc and L = Lc, assuming there only one line of

ciritcal in the (K,L) plane [8]. Then we must require K = K∗c = Kc and L = Lc = L∗c ,

we get for the T = Tc:

sinh 2K sinh 2L = 1 (A.11)

47

Appendix B

This appendix lists some properties of the elliptic functions and relating them to the

solution of the Square-lattice Ising model.

Properties of Elliptic Functions

When introducing the elliptic functions being periodic in two directions we heavily rely

on properties associated functions being doubly periodic.

Definition B.1. Any function f(u) satisfying

f(u + 2I) = (−1)sf(u)f(u + 2iI ′) = (−1)rf(u) (B.1)

where r, s are integers, is said to be doubly periodic.

The elliptic functions being depends two variables u and q which oftenly and in our

case is a real variable restricted to 0 < q < 1. We then the half period magnitudes I and

I ′ for the elliptic functions is given by

I =1

2π∞∏n=1

(1 + q2n−11− q2n−1

· 1− q2n

1 + q2n

)2(B.2)

I ′ = π−1I ln(q−1) (B.3)

where we can write q simply as:

q = exp(−πI ′/I) (B.4)

Our solutions of the Square-Lattice model is based on the functional relation of K and

L where we restrict them to a real constant k. When introducing the elliptic functions

for a parametrization of k the modulus k and its conjugate k′ takes the form

k = 4q1/2∞∏n=1

( 1 + q2n1 + q2n−1

)4k′ =

∞∏n=1

(1− q2n−11 + q2n−1

)4(B.5)

but most importantly it is then defined for 0 < k < 1.

A key property of the theta functions is their analyticity and periodicity mainly

H(u) = 0 when u = 2mI + 2inI ′ (B.6)

Θ(u) = 0 when u = 2mI + i(2n− 1)I ′ (B.7)

48

where n and m are integers.

Lastly the theta functions are related by:

H(u+ 2I) = −H(u) (B.8)H(u+ 2iI ′) = −q−1 exp(−πiu/I)H(u) (B.9)

and

Θ(u) = −iq1/2 exp(12πiu/I)H(u+ iI ′) (B.10)

49

Appendix C

In this Appendix we outline theorems which is vital to solving the Square-lattice Ising

model.

I am not going to present any proof to any of these theorems and shamelessly I have

taking these from the true hero of the Square-lattice Ising; Baxter Rodney. For more

about these theorems and their role in the Ising model read his brilliant book [8].

General Theorems

Liouville’s theorem. Every bounded entire function must be constant

Corollary C.1. If a function is doubly periodic (or anti-periodic) and is analytic inside

and on a period rectangle, then it is a constant.

Theorem C.2. If a function f(u) is doubly periodic (or anti-periodic) and meromorphic,

and has n poles per period rectangle,then it also has just n zeros per period rectangle.

Multiple poles or zeros of order r being counted r times)

Theorem C.3. If a function f(u) is meromorphic and satisfies the (anti-) periodicity

conditions and if f(u) has just n poles per period rectangle, at v1,..., vn, (counting a pole

of order r as r coincident simple poles), then

f(u) = Ceiλun∏j=1

[H(u − uj)]/H(u − vj)] (C.1)

where C, λ, v1, ..., vn are constants satisfying

u1 + ...+ un = v1 + ...+ vn + (r + 2m)I − i(s+ 2n)I ′ (C.2)

λ =1

2π(s+ 2n)/I (C.3)

Theorem C.4. Perron–Frobenius theorem A real square matrix with positive entries has

a unique largest real eigenvalue and that the corresponding eigenvector can be chosen to

have strictly positive components.

50